Understanding the constraints to find a $2\times 2$ non-zero matrix $A$ such that $A^2=0$ Using the rules for matrix multiplication, I have found four algebraic equations as "constraints" for getting every element in the resulting matrix to be zero. Assuming that the elements in the resulting matrix are $a, b, c$ and $d$ ($a, b$ are the elements in the first row and $c, d$ are the elements in the second). The equations are therefore;


*

*$a^2 + bc = 0$

*$ab + bd = 0$

*$ac + cd = 0$

*$d^2 + bc = 0$
I have found from the four equations that $a=-d$ and that $bc=-a^2$ hence $bc$ must be  a negative quantity thus $b$ and $c$ have opposite signs but that does not seem to be enough to guarantee that the resulting matrix is always zero. Now, I do not know how to think about this problem; 
How do I find more constraints from these equations? 
And how do I know that I have found all possible constraints?
Why is that when I square one equation, some how I get a new constraint? 
Shouldn't the four equations be enough to determine the exact conditions for $a, b, c$ and $d$?
 A: With
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, \tag 1$
we have
$A^2 = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a^2 + bc & (a + d)b \\ (a +d)c & d^2 + bc \end{bmatrix} = 0, \tag 2$
whence
$a^2 + bc = 0, \tag 3$
$d^2 + bc = 0, \tag 4$
$(a + d)b = 0, \tag 5$
$(a + d)c = 0; \tag 6$
suppose 
$a + d \ne 0; \tag 7$
then via (5) and (6),
$b = c = 0; \tag 8$
then via (3) and (4),
$a^2 = d^2 = 0 \Longrightarrow a = d = 0 \Longrightarrow a + d = 0 \Rightarrow \Leftarrow a + d \ne 0; \tag 9$
this contradiction rules out the case (7), so
$a + d = 0 \Longrightarrow d = -a; \tag{10}$
now if
$a = d = 0, \tag{11}$
then (3)-(4) imply, assuming $A \ne 0$, exactly one of 
$b = 0, \; c \ne 0, \tag{12}$
$b \ne 0, \; c = 0, \tag{13}$
holds.  Thus the solutions in the event that (11) binds are one of the forms
$A = \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix}, \;  \begin{bmatrix} 0 & 0 \\ c & 0 \end{bmatrix}; \tag{14}$
when
$d = -a \ne 0, \tag{15}$
we find that (3)-(4) imply
$b, c \ne 0 \tag{16}$
and
$c = -\dfrac{a^2}{b}; \tag{16}$
therefore
$A = \begin{bmatrix} a & b \\ -\dfrac{a^2}{b} & -a \end{bmatrix}. \tag{17}$
We see that (2) implies $A$ takes one of the three forms (14), (17); the former being comprised of two one-parameter families of matrices, and the latter comprised of one two-parameter family.
A: There's an easier way to do it. A $2\times2$ matrix $A$ such that $A^2=0$ must have the single eigenvalue $0$, so its trace and determinant must be zero, because the characteristic polynomial is $X^2-\operatorname{tr}(A)X+\det(A)$.
This yields, with your notation, $a+d=0$ and $ad-bc=0$, so $d=-a$ and $a^2+bc=0$.
Note that these conditions are sufficient, because the minimal polynomial is a divisor of the characteristic polynomial, so it is either $X$ (null matrix) or $X^2$ (nonzero nilpotent matrix).
Now you can classify the nonzero nilpotent matrices as those of the form
\begin{align}
&\begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix}
&&(b\ne0) \\[6px]
&\begin{bmatrix} 0 & 0 \\ c & 0 \end{bmatrix}
&&(c\ne0) \\[6px]
&\begin{bmatrix} a & b \\ -a^2/b & -a \end{bmatrix}
&&(a\ne0,b\ne0)
\end{align}
A: If $b=c=0$ then you get the matrIx of the form $$\begin{pmatrix} a & 0 \\ 0  & -a \\ \end{pmatrix}.$$
Now suppose that both of them are not simultaneously $0.$ WLOG let us assume that $b \neq 0.$ Then the matrix is of the form $$\begin{pmatrix} a & b \\ -\frac {a^2} {b} & -a \\ \end{pmatrix}.$$
So this kind of matrices have at most $2$ degrees of freedom.
